Tempered Distributions and the Fourier Transform

Home , Schwartz kernel theorem, Support (mathematics)

CHAPTER 1

Tempered distributions and the Fourier transform

Microlocal analysis is a geometric theory of distributions, or a theory of geometric distributions. Rather than study general distributions – which are like general continuous functions but worse – we consider more specific types of distributions which actually arise in the study of differential and integral equations. Distri- butions are usually defined by duality, starting from very “good” test functions; correspondingly a general distribution is everywhere “bad”. The conormal distributions we shall study implicitly for a long time, and eventually explicitly, are usually good, but like (other) people have a few interesting faults, i.e. singularities. These singularities are our principal target of study. Nevertheless we need the general framework of distribution theory to work in, so I will start with a brief introduction. This is designed either to remind you of what you already know or else to send you off to work it out. (As noted above, I suggest Friedlander’s little book [4] - there is also a newer edition with Joshi as coauthor) as a good introduction to distributions.) Volume 1 of Hörmander’streatise [8] has all that you would need; it is a good general reference. Proofs of some of the main theorems are outlined in the problems at the end of the chapter.

1.1. Schwartz test functions To fix matters at the beginning we shall work in the space of tempered distributions. These are defined by duality from the space of Schwartz functions, also called the space of test functions of rapid decrease. We can think of analysis as starting off from algebra, which gives us the polynomials. Thus in Rn we have the coordinate functions, x1, . . . , xn and the constant functions and then the polynomials are obtained by taking (finite) sums and products:

X α n (1.1) φ(x) = pαx , pα ∈ C, α ∈ N0 , α = (α1, . . . , αn), |α|≤k n α α1 αn Y αj where x = x1 . . . xn = xj and N0 = {0, 1, 2,... }. j=1 A general function φ : Rn −→ C is diﬀerentiable atx ¯ if there is a linear function n P `x¯(x) = c + (xj − x¯j)dj such that for every > 0 there exists δ > 0 such that j=1 ¯ (1.2) |φ(x) − `x¯(x)| ≤ |x − x¯| ∀ |x − x¯| < δ.

The coefficients dj are the partial derivative of φ at the pointx. ¯ Then, φ is said to be differentiable on Rn if it is differentiable at each pointx ¯ ∈ Rn; the partial derivatives are then also functions on Rn and φ is twice differentiable if the partial

13 14 1. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORM derivatives are differentiable. In general it is k times differentiable if its partial derivatives are k − 1 times differentiable. If φ is k times differentiable then, for eachx ¯ ∈ Rn, there is a polynomial of degree k, X |α| α pk(x;x ¯) = aαi (x − x¯) /α!, |α| = α1 + ··· + αn, |α|≤k such that for each > 0 there exists δ > 0 such that k (1.3) |φ(x) − pk(x, x¯)| ≤ |x − x¯| if |x − x¯| < δ. Then we set α (1.4) D φ(¯x) = aα. If φ is infinitely differentiable all the Dαφ are infinitely differentiable (hence continuous!) functions. Definition 1.1. The space of Schwartz test functions of rapid decrease consists n n of those φ : R −→ C such that for every α, β ∈ N0 (1.5) sup |xβDαφ(x)| < ∞; n x∈R it is denoted S(Rn). From (1.5) we construct norms on S(Rn): α β (1.6) kφkk = max sup |x D φ(x)|. |α|+|β|≤k n x∈R It is straightforward to check the conditions for a norm:

(1) kφkk ≥ 0, kφkk = 0 ⇐⇒ φ ≡ 0 (2) ktφkk = |t|kφkk, t ∈ C n (3) kφ + ψkk ≤ kφkk + kψkk ∀ φ, ψ ∈ S(R ). The topology on S(Rn) is given by the metric

X kφ − ψkk (1.7) d(φ, ψ) = 2−k . 1 + kφ − ψkk k See Problem 1.4.

Proposition 1.1. With the distance function (1.7), S(Rn) becomes a complete metric space (in fact it is a Fr´echetspace).

Of course one needs to check that S(Rn) is non-trivial; however one can easily see that 2 n (1.8) exp(−|x| ) ∈ S(R ). In fact there are lots of smooth functions of compact support and ∞ n n n (1.9) Cc (R ) = {u ∈ S(R ); u = 0 in |x| > R = R(u)} ⊂ S(R ) is dense. The two elementary operations of diﬀerentiation and coordinate multiplication give continuous linear operators: n n xj : S(R ) −→ S(R ) (1.10) n n Dj : S(R ) −→ S(R ). 1.2. LINEAR TRANSFORMATIONS 15

Other important operations we shall encounter include the exterior product, n m n+m S(R ) × S(R ) 3 (φ, ψ) 7→ φ ψ ∈ S(R ) (1.11) φ ψ(x, y) = φ(x)ψ(y). k n and pull-back or restriction. If R ⊂ R is identiﬁed as the subspace xj = 0, j > k, then the restriction map ∗ n k ∗ (1.12) πk : S(R ) −→ S(R ), πkf(y) = f(y1, . . . , yk, 0,..., 0) is continuous (and surjective).

1.2. Linear transformations

A linear transformation acts on Rn as a matrix (This is the standard action, n but it is potentially confusing since it means that for the basis elements ej ∈ R , n P Lej = Lkjek.) k=1 n n n X (1.13) L : R −→ R , (Lx)j = Ljkxk. k=1 The Lie group of invertible linear transformations, GL(n, R) is ﬁxed by several equivalent conditions L ∈ GL(n, R) ⇐⇒ det(L) 6= 0 −1 −1 n (1.14) ⇐⇒ ∃ L s.t. (L )Lx = x ∀ x ∈ R −1 n ⇐⇒ ∃ c > 0 s.t. c|x| ≤ |Lx| ≤ c |x| ∀ x ∈ R . Pull-back of functions is deﬁned by L∗φ(x) = φ(Lx) = (φ ◦ L)(x).

The chain rule for diﬀerentiation shows that if φ is diﬀferentiable then (So Dj transforms as a basis of Rn as it should, despite the factors of i.)

n ∗ X ∗ (1.15) DjL φ(x) = Djφ(Lx) = Lkj(Dkφ)(Lx) = L ((L∗Dj)φ)(x), k=1 n X L∗Dj = LkjDk. k=1 From this it follows that ∗ n n (1.16) L : S(R ) −→ S(R ) is an isomorphism for L ∈ GL(n, R). To characterize the action of L ∈ GL(n, R) on S0(Rn) consider, as usual, the distribution associated to L∗φ : Z (1.17) TL∗φ(ψ) = φ(Lx)ψ(x)dx n R Z −1 −1 −1 −1 ∗ = φ(y)ψ(L y)| det L| dy = Tφ(| det L| (L ) ψ). n R 16 1. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORM

Since the operator | det L|−1(L−1)∗ is an ismorphism of S(Rn) it follows that if we take the deﬁnition by duality

∗ −1 −1 ∗ 0 n n (1.18) L u(ψ) = u(| det L| (L ) ψ), u ∈ S (R ), ψ ∈ S(R ),L ∈ GL(n, R) ∗ 0 n 0 n =⇒ L : S (R ) −→ S (R ) is an isomorphism which extends (1.16) and satisﬁes (1.19) ∗ ∗ ∗ ∗ ∗ 0 n DjL u = L ((L∗Dj)u),L (xju) = (L xj)(L u), u ∈ S (R ),L ∈ GL(n, R), as in (1.15).

1.3. Tempered distributions As well as exterior multiplication (1.11) there is the even more obvious multiplication operation n n n S(R ) × S(R ) −→ S(R ) (1.20) (φ, ψ) 7→ φ(x)ψ(x) which turns S(Rn) into a commutative algebra without identity. There is also integration Z n (1.21) : S(R ) −→ C. Combining these gives a pairing, a bilinear map Z n n (1.22) S(R ) × S(R ) 3 (φ, ψ) 7−→ φ(x)ψ(x)dx. n R If we fix φ ∈ S(Rn) this defines a continuous linear map: Z n (1.23) Tφ : S(R ) 3 ψ 7−→ φ(x)ψ(x)dx. Continuity becomes the condition: n (1.24) ∃ k, Ck s.t. |Tφ(ψ)| ≤ Ckkψkk ∀ ψ ∈ S(R ). We generalize this by denoting by S0(Rn) the dual space, i.e. the space of all continuous linear functionals 0 n n u ∈ S (R ) ⇐⇒ u : S(R ) −→ C n ∃ k, Ck such that |u(ψ)| ≤ Ckkψkk ∀ ψ ∈ S(R ). Lemma 1.1. The map n 0 n (1.25) S(R ) 3 φ 7−→ Tφ ∈ S (R ) is an injection. n R 2 Proof. For any φ ∈ S(R ),Tφ(φ) = |φ(x)| dx, so Tφ = 0 implies φ ≡ 0. If we wish to consider a topology on S0(Rn) it will normally be the weak topology, that is the weakest topology with respect to which all the linear maps 0 n n (1.26) S (R ) 3 u 7−→ u(φ) ∈ C, φ ∈ S(R ) are continuous. This just means that it is given by the seminorms n (1.27) S(R ) 3 u 7−→ |u(φ)| ∈ R 1.4. TWO BIG THEOREMS 17 where φ ∈ S(Rn) is fixed but arbitrary. The sets 0 n (1.28) {u ∈ S (R ); |u(φj)| < j, φj ∈ Φ} form a basis of the neighbourhoods of 0 as Φ ⊂ S(Rn) runs over finite sets and the j are positive numbers.

Proposition 1.2. The continuous injection S(Rn) ,→ S0(Rn), given by (1.25), has dense range in the weak topology. See Problem 1.8 for the outline of a proof. The maps xi,Dj extend by continuity (and hence uniquely) to operators 0 n 0 n (1.29) xj,Dj : S (R ) −→ S (R ). n This is easily seen by deﬁning them by duality. Thus if φ ∈ S(R ) set DjTφ = TDj φ, then Z Z

(1.30) TDj φ(ψ) = Djφψ = − φDjψ, the integration by parts formula. The deﬁnitions 0 n n (1.31) Dju(ψ) = u(−Djψ), xju(ψ) = u(xjψ), u ∈ S (R ), ψ ∈ S(R ) satisfy all requirements, in that they give continuous maps (1.29) which extend the standard deﬁnitions on S(Rn).

1.4. Two big theorems The association, by (1.25), of a distribution to a function can be extended considerably. For example if u : Rn −→ C is a bounded and continuous function then Z (1.32) Tu(ψ) = u(x)ψ(x)dx still deﬁnes a distribution which vanishes if and only if u vanishes identically. Using n the operations (1.29) we conclude that for any α, β ∈ N0 β α 0 n n (1.33) x Dx u ∈ S (R ) if u : R −→ C is bounded and continuous. Conversely we have the Schwartz representation Theorem:

0 n n Theorem 1.1. For any u ∈ S (R ) there is a ﬁnite collection uαβ : R −→ C of bounded continuous functions, |α| + |β| ≤ k, such that

X β α (1.34) u = x Dx uαβ. |α|+|β|≤k Thus tempered distributions are just products of polynomials and derivatives of bounded continuous functions. This is important because it says that distributions are “not too bad”. The second important result (long considered very diﬃcult to prove, but there is a relatively straightforward proof using the Fourier transform) is the Schwartz kernel theorem. To show this we need to use the exterior product (1.11). If K ∈ S0(Rn+m) this allows us to deﬁne a linear map m 0 n (1.35) OK : S(R ) −→ S (R ) 18 1. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORM by Z (1.36) OK (ψ)(φ) = K · φ ψ dxdy.

Theorem 1.2. There is a 1-1 correspondence between continuous linear operators

m 0 n (1.37) A : S(R ) −→ S (R )

0 n+m and S (R ) given by A = OK . Brief outlines of the proofs of these two results can be found in Problems 1.15 and 1.16.

1.5. Examples

Amongst tempered distributions we think of S(Rn) as being the ‘trivial’ examples, since they are the test functions. One can say that the study of the singularities of tempered distributions amounts to the study of the quotient

0 n n (1.38) S (R )/S(R ) which could, reasonably, be called the space of tempered microfunctions. The sort of distributions we are interested in are those like the Dirac delta “function”

0 n (1.39) δ(x) ∈ S (R ), δ(φ) = φ(0). The deﬁnition here shows that δ is just the Schwartz kernel of the operator

n 0 (1.40) S(R ) 3 φ 7−→ φ(0) ∈ C = S(R ). This is precisely one reason it is interesting. More generally we can consider the maps

n α n (1.41) S(R ) 3 φ 7−→ D φ(0), α ∈ N0 . These have Schwartz kernels (−D)αδ since

(1.42) (−D)αδ(φ) = δ(Dαφ) = Dαφ(0).

If we write the relationship A = OK ←→ K as Z (1.43) (Aψ)(φ) = K(x, y)φ(x)ψ(y)dxdy then (1.42) becomes Z (1.44) Dαφ(0) = (−D)αδ(x)φ(x)dx.

More generally, if K(x, y) is the kernel of an operator A then the kernel of A · Dα α α α is (−D)y K(x, y) whereas the kernel of D ◦ A is Dx K(x, y). 1.6. TWO LITTLE LEMMAS 19

1.6. Two little lemmas Above, some of the basic properties of tempered distributions have been outlined. The main “raison d’être”for S0(Rn) is the Fourier transform which we proceed to discuss. We shall use the Fourier transform as an almost indispensable tool in the treatment of pseudodifferential operators. The description of differential operators, via their Schwartz kernels, using the Fourier transform is an essential motivation for the extension to pseudodifferential operators. Partly as simple exercises in the theory of distributions, and more significantly as preparation for the proof of the inversion formula for the Fourier transform we consider two lemmas. 0 n 0 n First recall that if u ∈ S (R ) then we have defined Dju ∈ S (R ) by

n (1.45) Dju(φ) = u(−Djφ) ∀ φ ∈ S(R ).

In this sense it is a “weak derivative”. Let us consider the simple question of the form of the solutions to

0 n (1.46) Dju = 0, u ∈ S (R ).

Let Ij be the integration operator:

n n−1 Ij : S(R ) −→ S(R ) (1.47) Z Ij(φ)(y1, . . . , yn−1) = φ(y1, . . . yj−1, x, yj, . . . yn−1)dx.

n n−1 Then if πj : R −→ R is the map πj(x) = (x1, . . . , xj−1, xj+1 . . . , xn), we deﬁne, for v ∈ S0(Rn−1),

∗ n (1.48) πj v(φ) = v(Ijφ) ∀ φ ∈ S(R ).

n n−1 ∗ It is clear from (1.47) that Ij : S(R ) −→ S(R ) is continuous and hence πj v ∈ S0(Rn) is well-deﬁned for each v ∈ S0(Rn−1).

∗ Lemma 1.2. The equation (1.46) holds if and only if u = πj v for some v ∈ S0(Rn−1).

n n Proof. If φ ∈ S(R ) and φ = Djψ with ψ ∈ S(R ) then Ijφ = Ij(Djψ) = 0. ∗ Thus if u = πj v then

∗ (1.49) u(−Djφ) = πj v(−Djφ) = v(Ij(−Djφ)) = 0.

∗ Thus u = πj v does always satisfy (1.46). Conversely suppose (1.46) holds. Choose ρ ∈ S(R) with the property Z (1.50) ρ(x)dx = 1.

Then each φ ∈ S(Rn) can be decomposed as

n (1.51) φ(x) = ρ(xj)Ijφ(x1, . . . , xj−1, xj+1, . . . xn) + Djψ, ψ ∈ S(R ). 20 1. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORM

Indeed this is just the statement

n n ζ ∈ S(R ),Ijζ = 0 =⇒ ψ(x) ∈ S(R ) where x Z j ψ(x) = ζ(x1, . . . , xj−1, t, xj+1, . . . , xn)dt −∞ x Z j = ζ(x1, . . . , xj−1, t, xj+1, . . . , xn)dt. ∞ Using (1.51) and (1.46) we have

(1.52) u(φ) = u (ρ(xj)Ijφ(x1, . . . , xj−1, xj+1, . . . xn)) . Thus if

n−1 (1.53) v(ψ) = u (ρ(xj)ψ(x1, . . . , xj−1, xj+1, . . . xn)) ∀ ψ ∈ S(R ) 0 n−1 ∗ then v ∈ S (R ) and u = πj v. This proves the lemma.

∗ Of course the notation u = πj v is much too heavy-handed. We just write u(x) = v(x1, . . . , xj−1, xj+1, . . . , xn) and regard ‘v as a distribution in one addi- tional variable’. The second, related, lemma is just a special case of a general result of Schwartz concerning the support of a distribution. The particular result is:

0 n Lemma 1.3. Suppose u ∈ S (R ) and xju = 0, j = 1, . . . n then u = cδ(x) for some constant c.

Proof. Again we use the deﬁnition of multiplication and a dual result for n 1 test functions. Namely, choose ρ ∈ S(R ) with ρ(x) = 1 in |x| < 2 , ρ(x) = 0 in |x| ≥ 3/4. Then any φ ∈ S(Rn) can be written n X n (1.54) φ = φ(0) · ρ(x) + xjψj(x), ψj ∈ S(R ). j=1 This in turn can be proved using Taylor’s formula as I proceed to show. Thus

n X ∞ (1.55) φ(x) = φ(0) + xjζj(x) in |x| ≤ 1, with ζj ∈ C . j=1 Then,

n X (1.56) ρ(x)φ(x) = φ(0)ρ(x) + xjρζj(x) j=1

n and ρζj ∈ S(R ). Thus it suﬃces to check (1.54) for (1 − ρ)φ, which vanishes identically near 0. Then ζ = |x|−2(1 − ρ)φ ∈ S(Rn) and so n 2 X (1.57) (1 − ρ)φ = |x| ζ = xj(xjζ) j=1 1.7. FOURIER TRANSFORM 21

ﬁnally gives (1.54) with ψj(x) = ρ(x)ζj(x) + xjζ(x). Having proved the existence of such a decomposition we see that if xju = 0 for all j then n X (1.58) u(φ) = u(φ(0)ρ(x)) + u(xjψj) = cφ(0), c = u(ρ(x)), j=1 i.e. u = cδ(x). 1.7. Fourier transform Our normalization of the Fourier transform will be Z (1.59) Fφ(ξ) = e−iξ·xφ(x)dx.

As you all know the inverse Fourier transform is given by Z (1.60) Gψ(x) = (2π)−n eix·ξψ(ξ) dξ.

Since it is so important here I will give a proof of this invertibility. First however, let us note some of the basic properties. Both F and G give continuous linear maps n n (1.61) F, G : S(R ) −→ S(R ). To see this observe first that the integrals in (1.59) and (1.60) are absolutely convergent: Z Z (1.62) |Fφ(ξ)| ≤ |φ(x)|dx ≤ (1 + |x|2)−n dx × sup (1 + |x|2)n|φ(x)|, n x∈R where we use the definition of S(Rn). In fact this shows that sup |Fφ| < ∞ if φ ∈ S(Rn). Formal differentiation under the integral sign gives an absolutely convergent integral: Z Z −ixξ −ix·ξ DjFφ(ξ) = Dξj e φ(x)dx = e (−xjφ)dx

2 n since sup(1 + |x| ) |xjφ| < ∞. Then it follows that DjFφ is also bounded, i.e. Fφ x is diﬀerentiable, and (1.7) holds. This argument can be extended to show that Fφ is C∞, (1.63) DαFφ(ξ) = F(−x)αφ. Similarly, starting from (1.59), we can use integration by parts to show that Z Z −ixξ −ix·ξ ξjFφ(ξ) = e ξjφ(x)dx = e (Djφ)(x)dx i.e. ξjFφ = F(Djφ). Combining this with (1.63) gives α β α β (1.64) ξ Dξ Fφ = F D · [(−x) φ] . α β n Since Dx ((−x) φ) ∈ S(R ) we conclude α β n (1.65) sup |ξ Dζ Fφ| < ∞ =⇒ Fφ ∈ S(R ). This map is continuous since α β 2 n α β sup |ξ Dξ Fφ| ≤ C · sup |(1 + |x| ) Dx [(−x) φ] x

=⇒ kFφkk ≤ Ckkφkk+2n, ∀ k. 22 1. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORM

The identity (1.64), written in the form

F(Djφ) = ξjFφ (1.66) F(xjφ) = −Dξj Fφ is already the key to the proof of invertibility:

Theorem 1.3. The Fourier transform gives an isomorphism F : S(Rn) ←→ S(Rn) with inverse G. Proof. We shall use the idea of the Schwartz kernel theorem. It is important not to use this theorem itself, since the Fourier transform is a key tool in the (simplest) proof of the kernel theorem. Thus we consider the composite map n n (1.67) G ◦ F : S(R ) −→ S(R ) and write down its kernel. Namely ZZZ K(φ) = (2π)−n eiy·ξ−ix·ξφ(y, x)dxdξdy (1.68) n n 0 2n ∀ φ ∈ S(Ry × Rx) =⇒ K ∈ S (R ). The integrals in (1.68) are iterated, i.e. should be performed in the order indicated. Notice that if ψ, ζ ∈ S(Rn) then indeed Z Z Z (1.69) (G·F(ψ))(ζ) = ζ(y)(2π)−n eiy·ξ e−ix·ξψ(x)dx dξ dy dξ dy

= K(ζ ψ) so K is the Schwartz kernel of G·F. The two identities (1.66) translate (with essentially the same proofs) to the conditions on K : ( (D + D )K(x, y) = 0 (1.70) xj yj j = 1, . . . , n. (xj − yj)K(x, y) = 0 Next we use the freedom to make linear changes of variables, setting 0 2n KL(x, z) = K(x, x − z),KL ∈ S (R ) (1.71) i.e. KL(φ) = K(ψ), ψ(x, y) = φ(x, x − y) where the notation will be explained later. Then (1.70) becomes

(1.72) Dxj KL(x, z) = 0 and zjKL(x, z) = 0 for j = 1, . . . n

This puts us in a position to apply the two little lemmas. The ﬁrst says KL(x, z) = f(z) for some f ∈ S0(Rn) and then the second says f(z) = cδ(z). Thus (1.73) K(x, y) = cδ(x − y) =⇒ G · F = c Id . It remains only to show that c = 1. That c 6= 0 is obvious (since F(δ) = 1). The easiest way to compute the constant is to use the integral identity ∞ Z −x2 1 (1.74) e dx = π 2

−∞ to show that1

1See Problem 1.9. 1.7. FOURIER TRANSFORM 23

−|x|2 n −|ξ|2/4 F(e ) = π 2 e −|ξ|2/4 − n −|x|2 (1.75) =⇒ G(e ) = π 2 e =⇒ G · F = Id .

Now (2π)nG is actually the adjoint of F : Z Z n n (1.76) φ(ζ)Fψ(ζ)dζ = (2π) (Gφ) · ψdx ∀ φ, ψ ∈ S(R ).

It follows that we can extend F to a map on tempered distributions 0 n 0 n F : S (R ) −→ S (R ) (1.77) n n Fu(φ) = u((2π) Gφ) ∀ φ ∈ S(R ) Then we conclude

Corollary 1.1. The Fourier transform extends by continuity to an isomorphism

0 n 0 n (1.78) F : S (R ) ←→ S (R ) with inverse G, satisfying the identities (1.66). Although I have not discussed Lebesgue integrability I assume familiarity with the basic Hilbert space

2 n L (R ) = Z n 2 u : R −→ C; f is measurable and |f(x)| dx < ∞ / ∼, n R f ∼ g ⇐⇒ f = g almost everywhere.

This also injects by the same integration map (1.102) with S(Rn) as a dense subset n 2 p n S(R ) ,→ L (R ) ,→ S(R ). Proposition 1.3. The Fourier transform extends by continuity from the dense subspace S(Rn) ⊂ L2(Rn), to an isomorphism 2 n 2 n F : L (R ) ←→ L (R ) 1 n satisfying kFukL2 = (2π) 2 kukL2 .

Proof. Given the density of S(Rn) in L2(Rn), this is also a consequence of (1.76), since setting φ = Fu, for u ∈ S(Rn), gives Parseval’s formula Z Z Fu(ζ)Fv(ζ) = (2π)n u(x)v(x)dx.

Setting v = u gives norm equality (which is Plancherel’s formula). An outline of the proof of the density statement is given in the problems below. 24 1. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORM

1.8. Differential operators The simplest examples of the Fourier transform of distributions are immediate consequences of the definition and (1.66). Thus (1.79) F(δ) = 1 as already noted and hence, from (1.66), α α n (1.80) F(D δ(x)) = ξ ∀ α ∈ N0 . Now, recall that the space of distributions with support the point 0 is just: 0 n X α (1.81) u ∈ S (R ); sup(u) ⊂ {0} = u = cαD δ . finite Thus we conclude that the Fourier transform gives an isomorphism 0 n (1.82) F : u ∈ S (R ); supp(u) ⊂ {0} ←→ C[ξ] = {polynomials in ξ}. Another way of looking at this same isomorphism is to consider partial differential operators with constant coefficients: n n P (D): S(R ) −→ S(R ) (1.83) X α P (D) = cαD . The identity becomes n (1.84) F(P (D)φ)(ξ) = P (ξ)F(φ)(ξ) ∀ φ ∈ S(R ) and indeed the same formula holds for all φ ∈ S0(Rn). Using the simpler notation uˆ(ξ) = Fu(ξ) this can be written

X α (1.85) P (\D)u(ξ) = P (ξ)û(ξ),P (ξ) = cαξ . The polynomial P is called the (full) characteristic polynomial of P (D); of course it determines P (D) uniquely. It is important for us to extend this formula to differential operators with variable coefficients. Using (1.59) and the inverse Fourier transform we get ZZ (1.86) P (D)u(x) = (2π)−n ei(x−y)·ξP (ξ)u(y)dydξ where again this is an iterated integral. In particular the inversion formula is just the case P (ξ) = 1. Consider the space ∞ n n α (1.87) C∞ (R ) = u : R −→ C; sup |D u(x)| < ∞ ∀ α x the space of C∞ function with all derivatives bounded on Rn. Of course n ∞ n (1.88) S(R ) ⊂ C∞ (R ) ∞ n ∞ n but C∞ (R ) is much bigger, in particular 1 ∈ C∞ (R ). Now by Leibniz’ formula X α (1.89) Dα(uv) = Dβu · Dα−βv β β≤α n ∞ n it follows that S(R ) is a module over C∞ (R ). That is, ∞ n n n (1.90) u ∈ C∞ (R ), φ ∈ S(R ) =⇒ uφ ∈ S(R ). 1.9. RADIAL COMPACTIFICATION 25

From this it follows that if

X α ∞ n (1.91) P (x, D) = pα(x)D , pα ∈ C∞ (R ) |α|≤m then P (x, D): S(Rn) −→ S(Rn). The formula (1.86) extends to Z (1.92) P (x, D)φ = (2π)−n ei(x−y)·ξP (x, ξ)φ(y)dydξ where again this is an iterated integral. Here

X α (1.93) P (x, ξ) = pα(x)ξ |α|≤m is the (full) characteristic polynomial of P.

1.9. Radial compactiﬁcation For later purposes, and general propaganda, consider the quadratic radial com- pactiﬁcation of Rn. The smooth map

n x n (1.94) QRC : R 3 x 7−→ 1 ∈ R (1 + |x|2) 2 is 1-1 and maps onto the interior of the unit ball, Bn = {|x| ≤ 1}. Consider the subspace

∞ n n n (1.95) C˙ (B ) = {u ∈ S(R ); supp(u) ⊂ B }. This is just the set of smooth functions on Rn which vanish outside the unit ball. Then the composite (‘pull-back’) map

∗ ∞ n n (1.96) QRC : C˙ (B ) 3 u 7−→ u ◦ QRC ∈ S(R ) is a topological isomorphism. A proof is indicated in the problems below. The dual space of C˙∞(Bn) is generally called the space of ‘extendible distributions’ on Bn – because they are all given by restricting elements of S0(Rn) to C˙∞(Bn). Thus QRC also identifies the tempered distributions on Rn with the extendible distributions on Bn. We shall see below that various spaces of functions on Rn take interesting forms when pulled back to Bn. I often find it useful to ‘bring infinity in’ in this way. Why is this the ‘quadratic’ radial compactification, and not just the radial compactification? There is a good reason which is discussed in the problems below. The actual radial compactification is a closely related map which identifies Euclidean space, Rn, with the interior of the upper half of the n-sphere in Rn+1 : n 1 x (1.97) RC : R 3 x 7−→ 1 , 1 (1 + |x|2) 2 (1 + |x|2) 2 n,1 0 n+1 2 0 2 ∈ S = {X = (X0,X ) ∈ R ; X0 ≥ 0,X0 + |X | = 1} Since the half-sphere is diffeomorphic to the ball (as compact manifolds with bound- ary) these two maps can be compared – they are not the same. However it is true that RC also identifies S(Rn) with C˙∞(Sn,1). 26 1. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORM

1.10. Problems

Problem 1.1. Suppose φ : Rn −→ C is a function such that for each point n x¯ ∈ R and each k ∈ N0 there exists a constant k > 0 and a polynomial pk(x;x ¯) (in x) for which

1 k+1 (1.98) |φ(x) − pk(x;x ¯)| ≤ |x − x¯| ∀ |x − x¯| ≤ k. k Does it follow that φ is infinitely differentiable – either prove this or give a counter- example. Problem 1.2. Show that the function u(x) = exp(x) cos[ex] ‘is’ a tempered distribution. Part of the question is making a precise statement as to what this means! Problem 1.3. Write out a careful (but not necessarily long) proof of the ‘easy’ direction of the Schwartz kernel theorem, that any K ∈ S0(Rn+m) defines a continuous linear operator m 0 n (1.99) OK : S(R ) −→ S (R ) [with respect to the weak topology on S0(Rn) and the metric topology on S(Rm)] by

(1.100) OK φ(ψ) = K(ψ φ). [Hint: Work out what the continuity estimate on the kernel, K, means when it is paired with an exterior product ψ φ.] Problem 1.4. Show that d in (1.6) is a metric on S(Rn). [Hint: If k · k is a norm on a vector space show that ku + vk kuk kvk ≤ + .] 1 + ku + vk 1 + kuk 1 + kvk n Problem 1.5. Show that a sequence φn in S(R ) is Cauchy, resp. converges to φ, with respect to the metric d in Problem 1.4 if and only if φn is Cauchy, resp. converges to φ, with respect to each of the norms k · kk. Problem 1.6. Show that a linear map F : S(Rn) −→ S(Rp) is continuous with respect to the metric topology given in Problem 1.4 if and only if for each k there exists N(k) ∈ N a constant Ck such that n kF φkk ≤ CkkφkN(k) ∀ φ ∈ S(R ). Give similar equivalent conditions for continuity of a linear map f : S(Rn) −→ C and for a bilinear map S(Rn) × S(Rp) −→ C. Problem 1.7. Check the continuity of (1.12). Problem 1.8. Prove Proposition 1.2. [Hint: It is only necessary to show that 0 n if u ∈ S (R ) is ﬁxed then for any of the open sets in (1.1), B, (with all the j > 0) n 0 0 there is an element φ ∈ S(R ) such that u − Tφ ∈ B. First show that if φ1, . . . φp is a basis for Φ then the set 0 0 n 0 (1.101) B = {v ∈ S (R ); |hv, φji| < δj is contained in B if the δj > 0 are chosen small enough. Taking the basis to be orthonormal, show that u − ψ ∈ B0 can be arranged for some ψ ∈ Φ.] 1.10. PROBLEMS 27

Problem 1.9. Compute the Fourier transform of exp(−|x|2) ∈ S(Rn). [Hint: The Fourier integral is a product of 1-dimensional integrals so it suffices to assume x ∈ . Then R Z Z −iξx −x2 −ξ2/4 −(x+ i ξ)2 e e dx = e e 2 dx. Interpret the integral as a contour integral and shift to the new contour where i x + 2 ξ is real.] Problem 1.10. Show that (1.23) makes sense for φ ∈ L2(Rn) (the space of (equivalence classes of) Lebesgue square-integrable functions and that the resulting map L2(Rn) −→ S0(Rn) is an injection. Problem 1.11. Suppose u ∈ L2(Rn) and that −n−1 2 n D1D2 ··· Dnu ∈ (1 + |x|) L (R ), where the derivatives are defined using Problem 1.10. Using repeated integration, show that u is necessarily a bounded continuous function. Conclude further that for u ∈ S0(Rn) α −n−1 2 n D u ∈ (1 + |x|) L (R ) ∀ |α| ≤ k + n (1.102) =⇒ Dαu is bounded and continuous for |α| ≤ k. [This is a weak form of the Sobolev embedding theorem.] Problem 1.12. The support of a (tempered) distribution can be defined in terms of the support of a test function. For φ ∈ S(Rn) the support, supp(φ), is the closure of the set of points at which it takes a non-zero value. For u ∈ S0(Rn) we define [ 0 n 0 (1.103) supp(u) = O{,O = {O ⊂ R open; supp(φ) ⊂ O =⇒ u(φ) = 0} . Show that the definitions for S(Rn) and S0(Rn) are consistent with the inclusion S(Rn) ⊂ S0(Rn). Prove that supp(δ) = {0}. Problem 1.13. For simplicity in R, i.e. with n = 1, prove Schwartz theorem concerning distributions with support the origin. Show that with respect to the norm k · kk the space (1.104) {φ ∈ S(R); φ(x) = 0 in |x| < , = (φ) > 0} is dense in k+1 (1.105) φ ∈ S(R); φ(x) = x ψ(x), ψ ∈ S(R) . Use this to show that 0 X ` (1.106) u ∈ S (R), supp(u) ⊂ {0} =⇒ u = c`Dxδ(x). `, finite Problem 1.14. Show that if P is a differential operator with coefficients in ∞ n C∞ (R ) then P is local in the sense that 0 n (1.107) supp(P u) ⊂ supp(u) ∀ u ∈ S (R ). The converse of this, for an operator P : S(Rn) −→ S(Rn) where (for simplicity) we assume n (1.108) supp(P u) ⊂ K ⊂ R 28 1. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORM for a fixed compact set K, is Peetre’s theorem. How would you try to prove this? (No full proof required.)

Problem 1.15. (Schwartz representation theorem) Show that, for any p ∈ R the map n 2 −p/2 −1 2 −p/2 n (1.109) Rp : S(R ) 3 φ 7−→ (1 + |x| ) F [(1 + |ξ| ) Fφ] ∈ S(R ) is an isomorphism and, using Problem 1.11 or otherwise, n (1.110) p ≥ n + 1 + k =⇒ kRpφkk ≤ CkkφkL2 , ∀ φ ∈ S(R ). t 0 n 0 n t Let Rp : S (R ) −→ S (R ) be the dual map (deﬁned by Tpu(φ) = u(Rpφ)). Show t 0 n that Rp is an isomorphism and that if u ∈ S (R ) satisﬁes n (1.111) |u(φ)| ≤ Ckφkk, ∀ φ ∈ S(R ) t 2 n then Rpu ∈ L (R ), if p ≥ n + 1 + k, in the sense that it is in the image of the map t in Problem 1.10. Using Problem 1.11 show that Rn+1(Rn+1+ku) is bounded and continuous and hence that X β α (1.112) u = x D uα,β |α|+|β|≤2n+2+k for some bounded continuous functions uα,β. Problem 1.16. (Schwartz kernel theorem.) Show that any continuous linear operator m 0 n T : S(Ry ) −→ S (Rx) extends to a continuous linear operator 2 −k/2 k m 2 −q/2 q n T : (1 + |y| ) H (Ry ) −→ (1 + |x| ) H (Rx) for some k and q. Deduce that the operator

2 (−n−1−q)/2 2 q/2 2 k/2 2 −k/2 T˜ = (1 + |Dx| ) (1 + |x| ) ◦ T ◦ (1 + |y| ) (1 + |D| ) : 2 m n L (R ) −→ C∞(R ) is continuous with values in the bounded continuous functions on Rn. Deduce that ˜ n 2 m 0 n+m T has Schwartz kernel in C∞(R ; L (R )) ⊂ S (R ) and hence that T itself has a tempered Schwartz kernel. Problem 1.17. Radial compactiﬁcation and symbols. Problem 1.18. Series of problems discussing double polyhomogeneous symbols.